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Exercise 4.6 · Q16

Q.If ∣x∣≤1|x| \le 1, then 2tan⁡−1x−sin⁡−12x1+x22\tan^{-1}x - \sin^{-1}\dfrac{2x}{1+x^2} is equal to

(1) tan⁡−1x\tan^{-1}x
(2) sin⁡−1x\sin^{-1}x
(3) 00
(4) π\pi
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The chapter's double-angle identity converting tan⁡−1\tan^{-1} to sin⁡−1\sin^{-1} applies exactly under the stated condition ∣x∣≤1|x|\le1, making the second term identical to the first.

Step 1. Recall the identity. For ∣x∣≤1|x|\le1: 2tan⁡−1x=sin⁡−12x1+x22\tan^{-1}x=\sin^{-1}\dfrac{2x}{1+x^2}. …

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